Since $\angle BLC = 135^\circ = \angle BLS$ we have $\angle CLS = 90^\circ$. We shall provide and prove several lemmas.

Lemma 1. Let $P$ be the intersection of the lines $LS, BC$ then $\angle PRB = 90^\circ$
Proof. Since the line $LC$ is tangent to the circumcircle $ABC$
$$\angle LCE = \angle B = \angle LEC$$
Therefore $LC = LE$. Similarly $KB = KF$ and $LF = BR$. Let $X$ be a point on the ray $CL$ such that $LF = LX$ such that $L$ is between $X, C$.
$$LE \cdot LF = LC \cdot LX$$
hence $EXFC$ is cyclic and $\angle BXC = 90^\circ$. Then
$$\frac{PC}{PB} = \frac{LC}{LX}; \\ LX = LF = BR; \\ \angle LCP = \angle A - 90^\circ = \angle RBP$$
Thus the triangles $LCP, RBP$ are similar and $\angle RBP = 90^\circ$. □

Let $Q$ be a point on line $BK$ such that $\angle BPQ = 90^\circ$ and $T$ be the second intersection of circumcircles $PLC, PQB$. It follows that

Lemma 2. Point $T$ lies on the circumcircle $ABC$
$$\begin{align*} \angle PTC &= \angle PLC = 90^\circ \\ \angle PTB &= \angle PQB = \angle A - 90^\circ \\ \angle CTB &= \angle PLC + \angle PTB = A \end{align*}$$
□

Note that $\angle LTP = \angle TCP = \angle TBS$, hence $SLTB$ is cyclic.
Now by our second lemma: $\angle RPB = \angle SPC$ and since $QPB = 90^\circ$, we would have $(BQ; RS) = -1$, further $QTB = 90^\circ$ therefore $TQ$ bisects $\angle RTS$.
Since $SLTB$ is cyclic $\angle STB = \angle SLB = 135^\circ$. Now it is easy to find that $\angle BTR = \angle RTQ = \angle QTS = 45^\circ$ thus the point $T$ lies on the circle with the diameter $RS$.

Lemma 3. Circumcircle $RTS$ is tangent to circumcircle $BTC$
Proof. Let $M$ be the midpoint of segment $BQ$.
Since $(BQ; RS) = -1$ we have
$$MR \cdot MS = MB^2 = MQ^2 = MT^2.$$
This concludes our proof.